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Bài làm:
Ta có: \(E=\frac{\sqrt{2x+2\sqrt{x^2-4}}}{\sqrt{x^2-4}+x+2}\)
\(E=\frac{\sqrt{\left(x+2\right)+2\sqrt{\left(x-2\right)\left(x+2\right)}+\left(x-2\right)}}{\sqrt{x^2-4}+x+2}\)
\(E=\frac{\sqrt{\left(\sqrt{x+2}+\sqrt{x-2}\right)^2}}{\sqrt{x^2-4}+x+2}\)
\(E=\frac{\sqrt{x+2}+\sqrt{x-2}}{\sqrt{x^2-4}+x+2}\)
Thay \(x=2\left(\sqrt{3}+1\right)\) vào thì giá trị của E là:
\(E=\frac{\sqrt{2\sqrt{3}+2+2}+\sqrt{2\sqrt{3}+2-2}}{\sqrt{\left(2\sqrt{3}+2\right)^2-4}+2\sqrt{3}+2+2}\)
\(E=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{2\sqrt{3}}}{\sqrt{12+4+8\sqrt{3}-4}+4+2\sqrt{3}}\)
\(E=\frac{\sqrt{3}+1+\sqrt{2\sqrt{3}}}{2\sqrt{3+2\sqrt{3}}+4+2\sqrt{3}}\)
a) \(\sqrt{\sqrt{2\sqrt{6}+6+2\sqrt{2}+2\sqrt{3}-\sqrt{5+2\sqrt{6}}}}\)
\(=\sqrt{1+\sqrt{2}+\sqrt{3}-\left(\sqrt{3}+\sqrt{2}\right)}=1\)
b) \(A=\sqrt{x^2-6x+9}-\dfrac{x^2-9}{\sqrt{9-6x+x^2}}\)
\(=\left|x-3\right|-\dfrac{\left(x-3\right)\left(x+3\right)}{\left|x-3\right|}\)
Th1: x-3 < 0
\(A=\left(3-x\right)-\dfrac{\left(x-3\right)\left(x+3\right)}{3-x}=3-x+x-3=0\)
Th2: x-3 > 0
\(A=x-3-\dfrac{\left(x-3\right)\left(x+3\right)}{x-3}=x-3-\left(x+3\right)=-6\)
c)
Đk: x >/ 1 \(B=\dfrac{\sqrt{x+\sqrt{4\left(x-1\right)}}-\sqrt{x-\sqrt{4\left(x-1\right)}}}{\sqrt{x^2-4\left(x-1\right)}}\cdot\left(\sqrt{x-1}-\dfrac{1}{\sqrt{x-1}}\right)\)
\(=\dfrac{\sqrt{x+2\sqrt{x-1}}-\sqrt{x-2\sqrt{x-1}}}{\sqrt{x^2-4\left(x-1\right)}}\cdot\dfrac{x-2}{\sqrt{x-1}}\)
\(=\dfrac{\sqrt{x-1}+1-\left|\sqrt{x-1}-1\right|}{\left|x-2\right|}\cdot\dfrac{x-2}{\sqrt{x-1}}\)
Th1: \(x-2\ge0\Leftrightarrow x\ge2\)
\(B=\dfrac{\sqrt{x-1}+1-\sqrt{x-1}+1}{x-2}\cdot\dfrac{x-2}{\sqrt{x-1}}=\dfrac{2}{\sqrt{x-1}}\)
Th2: \(x-2\le0\Leftrightarrow x\le2\)
kết hợp với đk, ta được: 1 \< x \< 2
\(=\dfrac{\sqrt{x-1}+1-\sqrt{x-1}-1}{2-x}\cdot\dfrac{x-2}{\sqrt{x-1}}=0\)
d) \(A=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}=\sqrt{x-2}+\sqrt{2}+\left|\sqrt{x-2}-\sqrt{2}\right|=\sqrt{x-2}+\sqrt{2}-\sqrt{x-2}+\sqrt{2}=2\sqrt{2}\)
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ĐKXĐ:...
\(E=\frac{\sqrt{x+2+2\sqrt{\left(x+2\right)\left(x-2\right)}+x-2}}{\sqrt{\left(x+2\right)\left(x-2\right)}+x+2}\)
\(=\frac{\sqrt{\left(\sqrt{x+2}+\sqrt{x-2}\right)^2}}{\sqrt{x+2}\left(\sqrt{x-2}+\sqrt{x+2}\right)}=\frac{1}{\sqrt{x+2}}\)
Câu sau đề đúng ko bạn? Thế này thì ko rút gọn được, tử số là \(\frac{3}{\sqrt{1+a}}+\sqrt{1-a}\) thì mới rút được
3) \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{4x-20}=4\)
\(\Leftrightarrow4x-20=16\)
\(\Leftrightarrow4x=36\)
\(\Leftrightarrow x=9\)
vậy ...
1)
\(A=\dfrac{\sqrt{x}-2}{x-4}=\dfrac{\sqrt{x}-2}{\left(\sqrt{x}\right)^2-2^2}\\ A=\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{1}{\sqrt{x}+2}\)
\(B=\dfrac{x^2-2x\sqrt{2}+2}{x^2-2}=\dfrac{x^2-2x\sqrt{2}+\left(\sqrt{2}\right)^2}{x^2-\sqrt{2}}\\ B=\dfrac{\left(x-\sqrt{2}\right)^2}{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}=\dfrac{\left(x-\sqrt{2}\right)}{\left(x+\sqrt{2}\right)}\)
\(C=\dfrac{x+\sqrt{5}}{x^2+2x\sqrt{5}+5}=\dfrac{x+\sqrt{5}}{x^2+2x\sqrt{5}+\left(\sqrt{5}\right)^2}\\ C=\dfrac{x+\sqrt{5}}{\left(x+\sqrt{5}\right)^2}=\dfrac{1}{x+\sqrt{5}}\)
\(D=\dfrac{\sqrt{a}-2a}{2\sqrt{a}-1}=\dfrac{\sqrt{a}\left(2\sqrt{a}-1\right)}{2\sqrt{a}-1}=\sqrt{a}\)
\(E=\dfrac{x^2-2}{x-\sqrt{2}}=\dfrac{x^2-\left(\sqrt{2}\right)^2}{x-\sqrt{2}}\\ E=\dfrac{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}{x-\sqrt{2}}=x+\sqrt{2}\)
\(F=\dfrac{\sqrt{x}-3}{x-9}=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}\right)^2-3^2}\\ F=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ F=\dfrac{1}{\sqrt{x}+3}\)
Lời giải:
a)
\(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}=\frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}+\frac{\sqrt{3}}{3}-\frac{2(3-\sqrt{3})}{(3+\sqrt{3})(3-\sqrt{3})}\)
\(=\frac{2-\sqrt{3}}{4-3}+\frac{\sqrt{3}}{3}-\frac{2(3-\sqrt{3})}{3^2-3}\)
\(=2-\sqrt{3}+\frac{\sqrt{3}}{3}-\frac{3-\sqrt{3}}{3}=\frac{6-3\sqrt{3}}{3}+\frac{2\sqrt{3}-3}{3}=\frac{3-\sqrt{3}}{3}\)
b)
\(\sqrt{x-3+2\sqrt{x-4}}=\sqrt{(x-4)+2\sqrt{x-4}+1}=\sqrt{(\sqrt{x-4}+1)^2}=|\sqrt{x-4}+1|=\sqrt{x-4}+1\)
c)
\(\sqrt{2x+4\sqrt{2x-4}}=\sqrt{(2x-4)+2.2\sqrt{2x-4}+2^2}\)
\(=\sqrt{(\sqrt{2x-4}+2)^2}=|\sqrt{2x-4}+2|=\sqrt{2x-4}+2\)
d)
\(\sqrt{x-\sqrt{2x-1}}=\frac{1}{\sqrt{2}}\sqrt{2x-2\sqrt{2x-1}}=\frac{1}{\sqrt{2}}\sqrt{(2x-1)-2\sqrt{2x-1}+1}\)
\(=\frac{1}{\sqrt{2}}\sqrt{(\sqrt{2x-1}-1)^2}=\frac{|\sqrt{2x-1}-1|}{\sqrt{2}}\)
e)
\(\sqrt{x+6\sqrt{x-9}}-\sqrt{x-9}=\sqrt{(x-9)+2.3\sqrt{x-9}+3^2}-\sqrt{x-9}\)
\(=\sqrt{(\sqrt{x-9}+3)^2}-\sqrt{x-9}=|\sqrt{x-9}+3|-\sqrt{x-9}\)
\(=\sqrt{x-9}+3-\sqrt{x-9}=3\)
Lời giải:
a)
\(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}=\frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}+\frac{\sqrt{3}}{3}-\frac{2(3-\sqrt{3})}{(3+\sqrt{3})(3-\sqrt{3})}\)
\(=\frac{2-\sqrt{3}}{4-3}+\frac{\sqrt{3}}{3}-\frac{2(3-\sqrt{3})}{3^2-3}\)
\(=2-\sqrt{3}+\frac{\sqrt{3}}{3}-\frac{3-\sqrt{3}}{3}=\frac{6-3\sqrt{3}}{3}+\frac{2\sqrt{3}-3}{3}=\frac{3-\sqrt{3}}{3}\)
b)
\(\sqrt{x-3+2\sqrt{x-4}}=\sqrt{(x-4)+2\sqrt{x-4}+1}=\sqrt{(\sqrt{x-4}+1)^2}=|\sqrt{x-4}+1|=\sqrt{x-4}+1\)
c)
\(\sqrt{2x+4\sqrt{2x-4}}=\sqrt{(2x-4)+2.2\sqrt{2x-4}+2^2}\)
\(=\sqrt{(\sqrt{2x-4}+2)^2}=|\sqrt{2x-4}+2|=\sqrt{2x-4}+2\)
d)
\(\sqrt{x-\sqrt{2x-1}}=\frac{1}{\sqrt{2}}\sqrt{2x-2\sqrt{2x-1}}=\frac{1}{\sqrt{2}}\sqrt{(2x-1)-2\sqrt{2x-1}+1}\)
\(=\frac{1}{\sqrt{2}}\sqrt{(\sqrt{2x-1}-1)^2}=\frac{|\sqrt{2x-1}-1|}{\sqrt{2}}\)
e)
\(\sqrt{x+6\sqrt{x-9}}-\sqrt{x-9}=\sqrt{(x-9)+2.3\sqrt{x-9}+3^2}-\sqrt{x-9}\)
\(=\sqrt{(\sqrt{x-9}+3)^2}-\sqrt{x-9}=|\sqrt{x-9}+3|-\sqrt{x-9}\)
\(=\sqrt{x-9}+3-\sqrt{x-9}=3\)
\(a,\sqrt{1-4a+4a^2}-2a\)
\(=\sqrt{\left(1-2a\right)^2}-2a\)
\(=1-2a-2a\)
\(=1-4a\)
\(b,x-2y-\sqrt{x^2-4xy+4y^2}\)
\(=x-2y-\sqrt{\left(x-2y\right)^2}\)
\(=x-2y-\left(x-2y\right)\)
\(=x-2y-x+2y\)
\(=0\)
\(c,x^2+\sqrt{x^4-8x^2+16}\)
\(=x^2+\sqrt{\left(x^2-4\right)^2}\)
\(=x^2+x^2-4\)
\(=2x^2-4\)
Các câu còn lại tương tự nha
\(a,\sqrt{1-4a+4a^2}-2a\)
\(=\sqrt{\left(1-2a\right)^2}-2a\)
\(=\left(1-2a\right)-2a\)
\(=1-4a\)
\(b,x-2y-\sqrt{x^2-4xy+4y^2}\)
\(=x-2y-\sqrt{\left(x-2y\right)^2}\)
\(=x-2y-\left(x-2y\right)\)
\(=x-2y-x+2y\)
\(=0\)
\(c,x^2+\sqrt{x^4-8x^2+16}\)
\(=x^2+\sqrt{\left(x^2-2^2\right)^2}\)
\(=x^2+\left(x^2-4\right)\)
\(=x^2+x^2-4\)
\(=2x^2-4\)
\(d,2x-1-\frac{\sqrt{x^2-10x+25}}{x-5}\)
\(=2x-1-\frac{\sqrt{\left(x-5\right)^2}}{x-5}\)
\(=2x-1-\frac{x-5}{x-5}\)
\(=2x-1-1\)
\(=2x-2\)
\(=2\left(x-1\right)\)
ĐK: x\(\ge\)2
\(E=\dfrac{\sqrt{x+2+2\sqrt{\left(x+2\right)\left(x-2\right)}+x-2}}{\sqrt{x^2-4}+x+2}\)
\(E=\dfrac{\sqrt{\left(\sqrt{x+2}+\sqrt{x-2}\right)^2}}{\sqrt{x^2-4}+x+2}\)
\(E=\dfrac{\left|\sqrt{x+2}+\sqrt{x-2}\right|}{\sqrt{x^2-4}+x+2}\)
\(E=\dfrac{\sqrt{x+2}+\sqrt{x-2}}{\left(x+2\right)+\sqrt{\left(x+2\right)\left(\sqrt{x-2}\right)}}\)
\(E=\dfrac{\sqrt{x+2}+\sqrt{x-2}}{\sqrt{x+2}\left(\sqrt{x+2}+\sqrt{x-2}\right)}\)
\(E=\dfrac{1}{\sqrt{x+2}}\)
Thế x=2(\(\sqrt{3}+1\))=\(2\sqrt{3}+2\) vào E:
=>\(E=\dfrac{1}{\sqrt{2\sqrt{3}+4}}\)
=>\(E=\dfrac{1}{\sqrt{3+2\sqrt{3}+1}}=\dfrac{1}{\sqrt{\left(\sqrt{3}+1\right)^2}}=\dfrac{1}{\sqrt{3}+1}\)